Answer:
The speed of the block after it has moved 2.92 m is 4.26 m/s
Explanation:
Given;
mass of the block, m = 2.97 kg
initial velocity of the block, u = 0
applied horizontal force, Fx = 12.9 N
Coefficient of static friction, μ = 0.126
The frictional force is given by;
[tex]F_s = \mu R\\\\F_s = \mu (mg)\\\\F_s = 0.126 (2.97*9.8)\\\\F_s = 3.667 \ N[/tex]
The net horizontal force is given by;
[tex]F_{net} = F_x - F_s\\\\F_{net} = 12.9 - 3.667\\\\F_{net} = 9.233 \ N[/tex]
The acceleration of the block is given by;
[tex]a = \frac{F_{net}}{m}\\\\a = \frac{9.233}{2.97}\\\\a = 3.11 \ m/s^2[/tex]
The speed of the block after it has moved 2.92 m is given by the following kinematic equation;
v² = u² + 2as
v² = 0 + 2as
v² = 2as
v = √2as
v = √(2 x 3.11 x 2.92)
v = 4.26 m/s
Therefore, the speed of the block after it has moved 2.92 m is 4.26 m/s
The speed of the block after it has moved 2.92 m is 4.26 m/s.
Given data:
The mass of block is, m = 2.97 kg.
The magnitude of horizontal force is, F = 12.9 N.
The coefficient of kinetic friction is, [tex]\mu = 0.126[/tex].
The magnitude of gravitational acceleration is, [tex]g = 9.8\;\rm m/s^{2}[/tex].
The distance moved by the block is, s = 2.92 m.
Let us find the net force acting on the block, which is equal to the difference between the horizontal force and frictional force. Then,
[tex]F_{net}= F - f\\\\F_{net}= F - (\mu \times m \times g)\\\\F_{net}= 12.9 - (0.126 \times 2.96 \times 9.8)\\\\F_{net}= 9.24 \;\rm N[/tex]
Now, use the Newton's second law of motion to obtain the magnitude of acceleration as,
[tex]F_{net} = m \times a\\\\9.24 = 2.97 \times a\\\\a =3.11 \;\rm m/s^{2}[/tex]
Finally, use the second kinematic equation of motion to find the final speed of block as,
[tex]v^{2}=u^{2}+2as\\\\v^{2}=0^{2}+(2 \times 3.11 \times 2.92)\\\\v = \sqrt{2 \times 3.11 \times 2.92}\\\\v =4.26 \;\rm m/s[/tex]
Thus, we can conclude that the speed of the block after it has moved 2.92 m is 4.26 m/s.
Learn more about the Newton's second law here:
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